Problem: Determine how many solutions exist for the system of equations. ${-12x-3y = -18}$ ${4x+y = 10}$
Convert both equations to slope-intercept form: ${-12x-3y = -18}$ $-12x{+12x} - 3y = -18{+12x}$ $-3y = -18+12x$ $y = 6-4x$ ${y = -4x+6}$ ${4x+y = 10}$ $4x{-4x} + y = 10{-4x}$ $y = 10-4x$ ${y = -4x+10}$ Just by looking at both equations in slope-intercept form, what can you determine? ${y = -4x+6}$ ${y = -4x+10}$ Both equations have the same slope with different y-intercepts. This means the equations are parallel. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ Parallel lines never intersect, thus there are NO SOLUTIONS.